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Question: If \(f\left( x \right) = {\left( {2011 + x} \right)^n}\), where \(x\) is a real variable and \(n\) i...

If f(x)=(2011+x)nf\left( x \right) = {\left( {2011 + x} \right)^n}, where xx is a real variable and nn is a positive integer, then value of f(0)+f′(0)+f′′2!+⋅⋅⋅⋅⋅f(n−1)(0)(n−1)f\left( 0 \right) + {f'}\left( 0 \right) + \dfrac{{{f{''}}}}{{2!}} + \cdot \cdot \cdot \cdot \cdot \dfrac{{{f^{\left( {n - 1} \right)}}\left( 0 \right)}}{{\left( {n - 1} \right)}} is
A. (2011)n{\left( {2011} \right)^n}
B. (2012)n{\left( {2012} \right)^n}
C. (2012)n−1{\left( {2012} \right)^n} - 1
D. n(2011)nn{\left( {2011} \right)^n}

Explanation

Solution

Find the value of the given condition f(0)+f′(0)+f′′2!+⋅⋅⋅⋅⋅f(n−1)(0)(n−1)f\left( 0 \right) + {f'}\left( 0 \right) + \dfrac{{{f{''}}}}{{2!}} + \cdot \cdot \cdot \cdot \cdot \dfrac{{{f^{\left( {n - 1} \right)}}\left( 0 \right)}}{{\left( {n - 1} \right)}} from binomial expansion by substituting the values in the function f(x)=(2011+x)nf\left( x \right) = {\left( {2011 + x} \right)^n} and simplify the expression to get the answer.

Complete step-by-step answer:
According to the question it is given that the function is,
f(x)=(2011+x)nf\left( x \right) = {\left( {2011 + x} \right)^n}
The value of f(0)f\left( 0 \right) can be defined by putting 00 in place of xx.
f(0)=(2011+0)nf\left( 0 \right) = {\left( {2011 + 0} \right)^n}
f(0)=(2011)nf\left( 0 \right) = {\left( {2011} \right)^n}……..(1)
Similarly, the value of f′(0){f'}\left( 0 \right) can be calculated by substituting 00 in place of xx.
f′(0)=nC12011n−1{f'}\left( 0 \right) = {}^n{C_1}{2011^{n - 1}}……….(2)
Proceed in the same way and find the value of all expressions.
f′′(0)2!=nC22011n−2\dfrac{{{f{''}}\left( 0 \right)}}{{2!}} = {}^n{C_2}{2011^{n - 2}} …………(3)
And,
fn−1(0)(n−1)!=nCn−12011\dfrac{{{f^{n - 1}}\left( 0 \right)}}{{\left( {n - 1} \right)!}} = {}^n{C_{n - 1}}2011
fn−1(0)(n−1)!=2011\dfrac{{{f^{n - 1}}\left( 0 \right)}}{{\left( {n - 1} \right)!}} = 2011……….(4)
Add the above all equations.
(2011)n+nC1(2011)n−1+nC2(2011)n−2+⋅⋅⋅⋅⋅nCn−12011{\left( {2011} \right)^n} + {}^n{C_1}{\left( {2011} \right)^{n - 1}} + {}^n{C_2}{\left( {2011} \right)^{n - 2}} + \cdot \cdot \cdot \cdot \cdot {}^n{C_{n - 1}}2011 …….(5)
Add and subtract nCn{}^n{C_n} in equation (5).
The above term (2011)n+nC1(2011)n−1+nC2(2011)n−2+⋅⋅⋅⋅⋅nCn−12011{\left( {2011} \right)^n} + {}^n{C_1}{\left( {2011} \right)^{n - 1}} + {}^n{C_2}{\left( {2011} \right)^{n - 2}} + \cdot \cdot \cdot \cdot \cdot {}^n{C_{n - 1}}2011 becomes,
(2011+1)n−1=(2012)n−1          (nCn=1){\left( {2011 + 1} \right)^n} - 1 = {\left( {2012} \right)^n} - 1\;\;\;\;\;\left( {{}^n{C_n} = 1} \right)

Hence, the option C is correct.

Note: Binomial expansion describes the algebraic expansion of the powers of a binomial. The binomial theorem specifies the expansion of any power. The larger the power is the harder it is to expand the expressions but binomial theorem makes it easier. Binomial means consisting of two terms
Expansion is defined as the act of getting bigger added onto something. Binomial expansion is very useful in finding the answers of any power. Binomial theorem is used in many fields.